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Let $ \Omega = \{z ∈ C : |z| < 2\}$ and $f$ be a function on $\Omega$ which is holomorphic at every point of $\Omega$ except $z = 1$ and at $z = 1 $ it has a simple pole. Suppose that $f(z) =\sum _{n=0}^\infty a_nz_n$, $|z| < 1$. Prove that $\lim_{n\to \infty }a_n = −c$, where $c$ is the residue of $f$ at $z=1$.

At every point $z_0;|z_0|<1$ ;$f$ has a Taylor series expansion $f(z) =\sum _{n=0}^\infty a_nz_n$; where $a_n=\dfrac{n!}{2\pi i}\int _\gamma \dfrac{f(z)}{(z-z_0)^{n+1}} dz$ where $z_0$ lies in the interior of $\gamma$.

How can I take the limit of $\lim_{n\to \infty }a_n$ from this expression ?Is it at all possible?

Please help.Only hints needed.

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    You know that $$-c = \lim_{z\to 1} (1-z)f(z),$$ don't you? – 2017-01-19
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    @DanielFischer; Since $z=1$ is a simple pole then $f(z)=\dfrac{\phi(z)}{z-1}$ where $\phi(1)\neq 0$ and $\phi$ is analytic at $z=1$ – 2017-01-19
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    $c=\text {Res} (f,1)=\lim_{z\to 1} (z-1)f(z)$ – 2017-01-19
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    How can I bring $\lim_ {n\to \infty }a_n$ into picture? – 2017-01-19
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    $c = \phi(1)$, use the power series expansion of $\phi$, like in the answer below. – 2017-01-19
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    I am not getting how is $c=\lim a_n$? – 2017-01-19

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Define the function $g(z) = (z-1)f(z)$, which is analytic on $\Omega$, because of the assumptions you have on $f$. Compute $$ g(z) = \sum_{n=0}^{\infty}a_{n}z^{n+1} - \sum_{n=0}^{\infty} a_n z^n = -a_0 + \sum_{n=1}^{\infty}(a_{n-1} - a_n)z^n$$ Since $g(1)$ exists, the serie $\Sigma_n (a_{n-1}-a_n) $ must converge ; thus $(a_n)$ converges and $\lim\limits_{n \to \infty} a_n = -c$

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    I get $g(1)=c\implies c=-a_0+\sum_{n=1}^\infty (a_{n-1}-a_n)$ – 2017-01-19
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    how do you get $\lim a_n=c$ – 2017-01-19
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    @BenStokes What is $-a_0 + \sum_{n = 1}^N (a_{n-1}-a_n)$? – 2017-01-19
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    Thank you very much,I got it ;@DanielFischer;@Marsan – 2017-01-19